Press, Teukolsly, Vetterling, Flannery - Numerical Recipes in C (523184), страница 32
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1980, Introduction to Numerical Analysis (New York: Springer-Verlag),§2.2. [1]Gear, C.W. 1971, Numerical Initial Value Problems in Ordinary Differential Equations (EnglewoodCliffs, NJ: Prentice-Hall), §6.2.Cuyt, A., and Wuytack, L. 1987, Nonlinear Methods in Numerical Analysis (Amsterdam: NorthHolland), Chapter 3.3.3 Cubic Spline InterpolationGiven a tabulated function yi = y(xi ), i = 1...N , focus attention on oneparticular interval, between xj and xj+1 . Linear interpolation in that interval givesthe interpolation formulay = Ayj + Byj+1(3.3.1)114Chapter 3.Interpolation and ExtrapolationwhereA≡xj+1 − xxj+1 − xjB ≡ 1−A =x − xjxj+1 − xj(3.3.2)Equations (3.3.1) and (3.3.2) are a special case of the general Lagrange interpolationformula (3.1.1).Since it is (piecewise) linear, equation (3.3.1) has zero second derivative inthe interior of each interval, and an undefined, or infinite, second derivative at theabscissas xj .
The goal of cubic spline interpolation is to get an interpolation formulathat is smooth in the first derivative, and continuous in the second derivative, bothwithin an interval and at its boundaries.Suppose, contrary to fact, that in addition to the tabulated values of yi , wealso have tabulated values for the function’s second derivatives, y , that is, a setof numbers yi . Then, within each interval, we can add to the right-hand side ofequation (3.3.1) a cubic polynomial whose second derivative varies linearly from aon the right. Doing so, we will have the desiredvalue yj on the left to a value yj+1continuous second derivative.
If we also construct the cubic polynomial to havezero values at xj and xj+1 , then adding it in will not spoil the agreement with thetabulated functional values yj and yj+1 at the endpoints xj and xj+1 .A little side calculation shows that there is only one way to arrange thisconstruction, namely replacing (3.3.1) byy = Ayj + Byj+1 + Cyj + Dyj+1(3.3.3)where A and B are defined in (3.3.2) andC≡1 3(A − A)(xj+1 − xj )26D≡1 3(B − B)(xj+1 − xj )26(3.3.4)Notice that the dependence on the independent variable x in equations (3.3.3) and(3.3.4) is entirely through the linear x-dependence of A and B, and (through A andB) the cubic x-dependence of C and D.We can readily check that y is in fact the second derivative of the newinterpolating polynomial. We take derivatives of equation (3.3.3) with respectto x, using the definitions of A, B, C, D to compute dA/dx, dB/dx, dC/dx, anddD/dx.
The result isyj+1 − yj3A2 − 13B 2 − 1dy=(xj+1 − xj )yj +(xj+1 − xj )yj+1−(3.3.5)dxxj+1 − xj66for the first derivative, andd2 y= Ayj + Byj+1dx2(3.3.6)for the second derivative. Since A = 1 at xj , A = 0 at xj+1 , while B is just theother way around, (3.3.6) shows that y is just the tabulated second derivative, andalso that the second derivative will be continuous across (e.g.) the boundary betweenthe two intervals (xj−1 , xj ) and (xj , xj+1 ).3.3 Cubic Spline Interpolation115The only problem now is that we supposed the yi ’s to be known, when, actually,they are not.
However, we have not yet required that the first derivative, computedfrom equation (3.3.5), be continuous across the boundary between two intervals. Thekey idea of a cubic spline is to require this continuity and to use it to get equationsfor the second derivatives yi .The required equations are obtained by setting equation (3.3.5) evaluated forx = xj in the interval (xj−1 , xj ) equal to the same equation evaluated for x = xj butin the interval (xj , xj+1). With some rearrangement, this gives (for j = 2, . .
. , N −1)xj+1 − xj−1 xj+1 − xj yj+1 − yjyj − yj−1xj − xj−1 yj−1 +yj +yj+1 =−636xj+1 − xjxj − xj−1(3.3.7)These are N − 2 linear equations in the N unknowns yi , i = 1, . . . , N . Thereforethere is a two-parameter family of possible solutions.For a unique solution, we need to specify two further conditions, typically takenas boundary conditions at x1 and xN . The most common ways of doing this are eitherequal to zero, giving the so-called natural• set one or both of y1 and yNcubic spline, which has zero second derivative on one or both of itsboundaries, or• set either of y1 and yNto values calculated from equation (3.3.5) so asto make the first derivative of the interpolating function have a specifiedvalue on either or both boundaries.One reason that cubic splines are especially practical is that the set of equations(3.3.7), along with the two additional boundary conditions, are not only linear, butalso tridiagonal.
Each yj is coupled only to its nearest neighbors at j ± 1. Therefore,the equations can be solved in O(N ) operations by the tridiagonal algorithm (§2.4).That algorithm is concise enough to build right into the spline calculational routine.This makes the routine not completely transparent as an implementation of (3.3.7),so we encourage you to study it carefully, comparing with tridag (§2.4).
Arraysare assumed to be unit-offset. If you have zero-offset arrays, see §1.2.#include "nrutil.h"void spline(float x[], float y[], int n, float yp1, float ypn, float y2[])Given arrays x[1..n] and y[1..n] containing a tabulated function, i.e., yi = f (xi ), withx1 < x2 < . . . < xN , and given values yp1 and ypn for the first derivative of the interpolatingfunction at points 1 and n, respectively, this routine returns an array y2[1..n] that containsthe second derivatives of the interpolating function at the tabulated points xi .
If yp1 and/orypn are equal to 1 × 1030 or larger, the routine is signaled to set the corresponding boundarycondition for a natural spline, with zero second derivative on that boundary.{int i,k;float p,qn,sig,un,*u;u=vector(1,n-1);if (yp1 > 0.99e30)The lower boundary condition is set either to be “naty2[1]=u[1]=0.0;ural”else {or else to have a specified first derivative.y2[1] = -0.5;u[1]=(3.0/(x[2]-x[1]))*((y[2]-y[1])/(x[2]-x[1])-yp1);}116Chapter 3.Interpolation and Extrapolationfor (i=2;i<=n-1;i++) {This is the decomposition loop of the tridiagonal alsig=(x[i]-x[i-1])/(x[i+1]-x[i-1]);gorithm. y2 and u are used for temp=sig*y2[i-1]+2.0;porary storage of the decomposedy2[i]=(sig-1.0)/p;factors.u[i]=(y[i+1]-y[i])/(x[i+1]-x[i]) - (y[i]-y[i-1])/(x[i]-x[i-1]);u[i]=(6.0*u[i]/(x[i+1]-x[i-1])-sig*u[i-1])/p;}if (ypn > 0.99e30)The upper boundary condition is set either to beqn=un=0.0;“natural”else {or else to have a specified first derivative.qn=0.5;un=(3.0/(x[n]-x[n-1]))*(ypn-(y[n]-y[n-1])/(x[n]-x[n-1]));}y2[n]=(un-qn*u[n-1])/(qn*y2[n-1]+1.0);for (k=n-1;k>=1;k--)This is the backsubstitution loop of the tridiagonaly2[k]=y2[k]*y2[k+1]+u[k];algorithm.free_vector(u,1,n-1);}It is important to understand that the program spline is called only once toprocess an entire tabulated function in arrays xi and yi .
Once this has been done,values of the interpolated function for any value of x are obtained by calls (as manyas desired) to a separate routine splint (for “spline interpolation”):void splint(float xa[], float ya[], float y2a[], int n, float x, float *y)Given the arrays xa[1..n] and ya[1..n], which tabulate a function (with the xai ’s in order),and given the array y2a[1..n], which is the output from spline above, and given a value ofx, this routine returns a cubic-spline interpolated value y.{void nrerror(char error_text[]);int klo,khi,k;float h,b,a;klo=1;We will find the right place in the table by means ofkhi=n;bisection. This is optimal if sequential calls to thiswhile (khi-klo > 1) {routine are at random values of x.
If sequential callsk=(khi+klo) >> 1;are in order, and closely spaced, one would do betterif (xa[k] > x) khi=k;to store previous values of klo and khi and test ifelse klo=k;they remain appropriate on the next call.}klo and khi now bracket the input value of x.h=xa[khi]-xa[klo];if (h == 0.0) nrerror("Bad xa input to routine splint"); The xa’s must be disa=(xa[khi]-x)/h;tinct.b=(x-xa[klo])/h;Cubic spline polynomial is now evaluated.*y=a*ya[klo]+b*ya[khi]+((a*a*a-a)*y2a[klo]+(b*b*b-b)*y2a[khi])*(h*h)/6.0;}CITED REFERENCES AND FURTHER READING:De Boor, C. 1978, A Practical Guide to Splines (New York: Springer-Verlag).Forsythe, G.E., Malcolm, M.A., and Moler, C.B. 1977, Computer Methods for MathematicalComputations (Englewood Cliffs, NJ: Prentice-Hall), §§4.4–4.5.Stoer, J., and Bulirsch, R.
1980, Introduction to Numerical Analysis (New York: Springer-Verlag),§2.4.Ralston, A., and Rabinowitz, P. 1978, A First Course in Numerical Analysis, 2nd ed. (New York:McGraw-Hill), §3.8.3.4 How to Search an Ordered Table1173.4 How to Search an Ordered TableSuppose that you have decided to use some particular interpolation scheme,such as fourth-order polynomial interpolation, to compute a function f(x) from aset of tabulated xi ’s and fi ’s.
Then you will need a fast way of finding your placein the table of xi ’s, given some particular value x at which the function evaluationis desired. This problem is not properly one of numerical analysis, but it occurs sooften in practice that it would be negligent of us to ignore it.Formally, the problem is this: Given an array of abscissas xx[j], j=1, 2, . . . ,n,with the elements either monotonically increasing or monotonically decreasing, andgiven a number x, find an integer j such that x lies between xx[j] and xx[j+1].For this task, let us define fictitious array elements xx[0] and xx[n+1] equal toplus or minus infinity (in whichever order is consistent with the monotonicity of thetable).
Then j will always be between 0 and n, inclusive; a value of 0 indicates“off-scale” at one end of the table, n indicates off-scale at the other end.In most cases, when all is said and done, it is hard to do better than bisection,which will find the right place in the table in about log2 n tries. We already did usebisection in the spline evaluation routine splint of the preceding section, so youmight glance back at that. Standing by itself, a bisection routine looks like this:void locate(float xx[], unsigned long n, float x, unsigned long *j)Given an array xx[1..n], and given a value x, returns a value j such that x is between xx[j]and xx[j+1].
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